For each positive real number λ , let A λ be the set of all natural numbers n such that | sin ⁡ ( n + 1 - sin ⁡ n ) | < λ . Let A λ c be the complement of A λ in the set of all natural numbers. Then-

A 1 2 c , A 1 3 c , A 2 5 c are all finite sets

A 1 2 , A 1 3 , A 2 5 are all finite sets

A 1 3 is a finite set but A 1 2 , A 2 5 are infinite sets

A 1 3 , A 2 5 are finite sets and A 1 2 is an infinite set

For each positive real number λ , let A λ be the set of all natural numbers n such that | sin ⁡ ( n + 1 - sin ⁡ n ) | < λ . Let A λ c be the complement of A λ in the set of all natural numbers. Then-

A 1 2 c , A 1 3 c , A 2 5 c are all finite sets

Let f : R → R be a function such that lim x → ∞ ⁡ f x = M > 0 . Then which of the following is false ?

lim x → ∞ ⁡ x sin ⁡ e - x f x = M

lim x → ∞ ⁡ x sin ⁡ 1 x f x = M

lim x → ∞ ⁡ sin ⁡ f x = sin ⁡ M

lim x → ∞ ⁡ sin ⁡ x x ⋅ f x = 0

HARD

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For each positive real number $λ,$ let $A_{λ}$ be the set of all natural numbers $n$ such that $| \sin (\sqrt{n + 1} - \sin n) | < λ .$ Let $A_{λ}^{c}$ be the complement of $A_{λ}$ in the set of all natural numbers. Then-

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Important Questions on Limits

MEDIUM

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

\lim_{x \to 1^{-}} \frac{\sqrt{π} - \sqrt{2 \sin^{- 1} x}}{\sqrt{1 - x}}

is equal to

EASY

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

\lim_{x \to \infty} (\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1})=

EASY

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

\lim_{x \to \infty} \frac{3 x^{3} + 2 x^{2} - 7 x + 9}{4 x^{3} + 9 x - 2}

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

f

is differentiable at

x = 1

,

then

\lim_{x \to 1} \frac{x^{2} f (1) - f (x)}{x - 1}

MEDIUM

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

\lim_{x \to \infty} (\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x})

is equal to

EASY

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

f (x) = \sqrt{\frac{x - \sin x}{x + \cos^{2} x}}

,

then

\lim_{x \to \infty} f (x)

is equal to

EASY

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

\lim_{x \to 0} (\frac{\int_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{2}}) =

MEDIUM

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

The value of the

\lim_{x \to 0} {(\frac{x}{\sin x})}^{6 / x^{2}}

is-

EASY

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

\lim_{x \to 0} \frac{{(27 + x)}^{\frac{1}{3}} - 3}{9 - {(27 + x)}^{\frac{2}{3}}}

equals

HARD

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

The limit

\underset{x \to \infty}{l i m} x^{2} \int_{0}^{x} e^{t^{3} - x^{3}} d t

equals

HARD

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

Let

f : R \to R

be a positive increasing function with

\lim_{x \to \infty} \frac{f (3 x)}{f (x)} = 1

. Then,

\lim_{x \to \infty} \frac{f (2 x)}{f (x)}

is equal to

HARD

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

[\cdot]

denote the greatest integer function then

\lim_{n \to \infty} \frac{[x] + [2 x] + \dots . + [n x]}{n^{2}}

is -

MEDIUM

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

\lim_{x \to 3} \frac{\sqrt{3 x} - 3}{\sqrt{2 x - 4} - \sqrt{2}}

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

\lim_{x \to 0} \frac{(\sqrt{1 + 2 x}) - 1}{x} =

MEDIUM

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

Let

a > 0

be a real number. Then the limit

\lim_{x \to 2} \frac{a^{x} + a^{3 - x} - (a^{2} + a)}{a^{3 - x} - a^{\frac{x}{2}}}

is-

MEDIUM

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

f (x)

is continuous and

f (\frac{9}{2}) = \frac{2}{9},

then

\lim_{x \to 0} f (\frac{1 - \cos 3 x}{x^{2}})

equals to

HARD

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

Let

f : R \to R

be a function such that

\lim_{x \to \infty} f (x) = M > 0

. Then which of the following is false?

MEDIUM

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

$\lim_{x \to \frac{π}{2}} \frac{\cot x - \cos x}{{(π - 2 x)}^{3}}$ equals

HARD

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

The value of the $\lim_{x \to - \infty} (\sqrt{4 x^{2} - x} + 2 x)$ is

EASY

Mathematics>Differential Calculus>Limits>Methods of Evaluation of Limits

$\underset{x \to \frac{π}{4}}{l i m} \frac{c o t^{3} x - t a n x}{c o s (x + \frac{π}{4})}$ is

For each positive real number λ, let Aλ be the set of all natural numbers n such that |sin⁡(n+1-sin⁡n)|<λ. Let Aλc be the complement of Aλ in the set of all natural numbers. Then-

Important Questions on Limits

Learn and Prepare for any exam you want !

For each positive real number $λ,$ let $A_{λ}$ be the set of all natural numbers $n$ such that $| \sin (\sqrt{n + 1} - \sin n) | < λ .$ Let $A_{λ}^{c}$ be the complement of $A_{λ}$ in the set of all natural numbers. Then-